Distribution of points on a rectangle Let $R$ be a rectangular region with sides $3$ and $4$. It is easy to show that for any $7$ points on $R$, there exists at least $2$ of them, namely $\{A,B\}$, with $d(A,B)\leq \sqrt{5}$. Just divide $R$ into six small rectangles with sides $2$ and $1$ and so at least one such rectangle must contain $2$ points from the seven ones. Thus the result follows.
Here is the question:  What about six points? 
I believe that the same is true. How do I prove my belief?

ps: I don't want to find such $6$ points. I'd like to show it for any set with $6$ points.
 A: For a different approach, also start by observing that any rectangular area with sides 2 and 1 can contain at most one point if any two points are to be more than $\sqrt{5}$ apart. Now note that 2-by-1 rectangles can be placed horizontally (blue in illustration below) or vertically (green).  So we can...
Split the 4-by-3 rectangle into 12 unit size squares, color like a chessboard (black and white such that no squares sharing an edge have the same color). 
Consider a pair of neighboring squares: If they contain (at least) two points, these have a distance $\leq \sqrt{5}$. If they contain no point, there must be another pair of adjacent squares with two points (pigeonhole principle, five 2-by-1 rectangles/square pairs left for six points), whose distance is then $\leq \sqrt{5}$. If each black/white pair of adjacent squares contains exactly one point, it follows that the points must be either all on black or all on white squares (there are exactly 6 of each).

Then, in either case there must be a "diagonal" of three squares with one point each (***** or x in illustration). Now imagine the middle square split along the other diagonal. Now for each triangle, every point in or on it is no farther than $\sqrt{5}$ from any point in the respective closer, diagonally "neighboring" square! (In other words, w.r.t. to the illustration, if the point in the middle square is closer to the top left, it is less than $\sqrt{5}$ from the point in the top left, otherwise it is within $\sqrt{5}$ distance of the  point in the lower right square.)
A: The problem and solution are in Jiří Herman, Radan Kučera, Jaromír Šimša, Counting and Configurations: Problems in Combinatorics, Arithmetic, and Geometry, page 272. Let the rectangle have corners $(0,0),(0,3),(4,0),(4,3)$. Draw line segments joining $(0,2)$ to $(1,1)$ to $(2,2)$ to $(3,1)$ to $(4,2)$, also $(1,1)$ to $(1,0)$, $(2,2)$ to $(2,3)$, and $(3,1)$ to $(3,0)$. This splits the rectangle into $5$ pieces, and it's not hard to show two points in the same piece must be within $\sqrt5$. 

A picture to illustrate the solution. 
 
A: Here is a variation from
a proof-almost-without-words at Puzzling SE.
This relies on the fact that all interior points of a
Reuleaux triangle
are no further apart than
the so-called triangle’s “width,”
just as with a circle and its diameter.


Note that 5 circles can’t quite fill the bill.


